Lecture3 (Chap 3)

7/28/2019 Lecture3 (Chap 3)

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Topic 3 : ELECTRIC CIRCUIT

Electromotive Force and Current

Ohms law and Resistivity

Electric Power

Alternating Current

Series and Parallel Wiring

Kirchhoffs Rules

The Measurement of Current and Voltage

Capacitors in Series and Parallel

RC Circuits

Safety and Physiological Effects of Current


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20.1Electrom ot ive Force and Current

In an electric circuit, an energy source and an energy consuming device

are connected by conducting wires through which electric charges move.


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Within a battery, a chemical reaction occurs that transfers electrons from

one terminal to another terminal.

The maximum potential difference across the terminals is called the

electrom otive force (emf).


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The electr ic c urrentis the amount of charge per unit time that passes

through a surface that is perpendicular to the motion of the charges.

tqI

One coulomb per second equals one ampere(A).


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If the charges move around the circuit in the same direction at all times,

the current is said to be direct current(dc).

If the charges move first one way and then the opposite way, the current issaid to be alternat ing c urrent(ac).


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Example 1A Pocket Calculator

The current in a 3.0 V battery of a pocket calculator is 0.17 mA. In one hour

of operation, (a) how much charge flows in the circuit and (b) how much energydoes the battery deliver to the calculator circuit?

(a)

(b)

C61.0s3600A1017.0 3 tIq

J8.1V0.3C61.0Charge

Energy

ChargeEnergy


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Conventional currentis the hypothetical flow of positive charges that would

have the same effect in the circuit as the movement of negative charges that

actually does occur.


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20.2Ohms Law

The resistance(R) is defined as theratio of the voltage V applied across

a piece of material to the current I through

the material.


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20.2Ohms Law

OHMS LAW

The ratio V/Iis a constant, where Vis thevoltage applied across a piece of material

and Iis the current through the material:

SI Unit o f Resistance:volt/ampere (V/A) = ohm ()

IRVRI

V orconstant


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20.2Ohms Law

To the extent that a wire or an electrical

device offers resistance to electrical flow,it is called a resistor .


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20.2Ohms Law

Example 2A Flashlight

The filament in a light bulb is a resistor in the form

of a thin piece of wire. The wire becomes hot enoughto emit light because of the current in it. The flashlight

uses two 1.5-V batteries to provide a current of

0.40 A in the filament. Determine the resistance of

the glowing filament.

5.7A0.40

V0.3

I

VR


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Exercise 1

A 3.5-A current is maintained in a simple

circuit with a total resistance of 1500 .

What net charge passes through any point

in the circuit during a thirty secondinterval?

Answer: 105 C


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20.3Resistance and Resist iv i ty

For a wide range of materials, the resistance

of a piece of material of length L and cross-

sectional areaA is

A

LR

resistivity in units of ohmmeter


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A

LR

20 3 R i t d R i t i i t


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Example 3Longer Extension Cords

The instructions for an electric lawn mower suggest that a 20-gauge extension

cord can be used for distances up to 35 m, but a thicker 16-gauge cord shouldbe used for longer distances. The cross sectional area of a 20-gauge wire is

5.2x10-7m2, while that of a 16-gauge wire is 13x10-7m2. Determine the

resistance of (a) 35 m of 20-gauge copper wire and (b) 75 m of 16-gauge

copper wire.

2.1m105.2

m35m1072.127-

8

A

LR (a)

(b)

99.0m1013

m75m1072.127-

8

A

LR


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Exercise 2

Determine the length of a copper wire that

has a resistance of 0.172 and cross-

sectional area of 7.85 105 m2. The

resistivity of copper is 1.72 108 . m.

Answer: 785m



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Impedance Plethysmography.

calf

2

calf V

L

LV

L

A

LR



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oo TT 1

temperature coefficientof resistivity

oo TTRR 1

20 4 Electr ic Power


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20.4Electr ic Power

Suppose some charge emerges from a battery and the potential difference

between the battery terminals is V.

IVV

t

q

t

VqP

energy

power

time



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20.4Electr ic Power

IVP

ELECTRIC POWER

When there is current in a circuit as a result of a voltage, the electric

power delivered to the circuit is:

SI Unit o f Power:watt (W)

Many electrical devices are essentially resistors:

RIIRIP 2

R

VV

R

VP

2


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20.4Electr ic Power

(a)

(b)

W2.1V0.3A40.0 IVP

J100.4s330W2.1 2 PtE


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Exercise 3

A 4-A current is maintained in a simple

circuit with a total resistance of 2 . How

much energy is delivered in forty five

seconds?

Answer: 1440J

20 5 Alternat ing Current


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20.5Alternat ing Current

In an AC circuit, the charge flow reverses direction periodically.



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ftVV o 2sin



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In circuits that contain only resistance, the current reverses direction each

time the polarity of the generator reverses.

f tIf tRV

R

V

I oo

2sin2sin

peak current



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ftVIIVP oo 2sin2

ftII o 2sin ftVV o 2sin

20.5 Al ternat ing Current


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rmsrms

222

VIVIVI

P oooo

20.5 Al ternat ing Current


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RIV rmsrms

rmsrmsIVP

RIP 2rms

RVP

2

rm s



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g

Example 6Electrical Power Sent to a

Loudspeaker

A stereo receiver applies a peak voltage of

34 V to a speaker. The speaker behaves

approximately as if it had a resistance of 8.0 .

Determine (a) the rms voltage, (b) the rms

current, and (c) the average power for this

circuit.



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g

(a)

(b)

(c)

V242

V34

2rm s

oVV

A3.0

0.8

V24rmsrms

R

VI

W72V24A3.0rm srms VIP


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Exercise 4

Calculate the resistance and the peak

current in a 1000W hair dryer connected to

a 120-V line.

Io = 11.8 A

R = 14.4

20.6Series Wiring


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g

There are many circuits in which more than one device is connected to

a voltage source.

Series w ir ing means th at the devices are con nected in such a way

that there is the same electr ic current throu gh each device.

20.6Series Wiring


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SIRRRIIRIRVVV 212121

321 RRRRSSeries resisto rs

20.6Series Wiring


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20.6Series Wiring


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Examp le 9Resistors in a Series Circuit

A 6.00 resistor and a 3.00 resistor are connected in series with

a 12.0 V battery. Assuming the battery contributes no resistance to

the circuit, find (a) the current, (b) the power dissipated in each resistor,

and (c) the total power delivered to the resistors by the battery.

20.6Series Wiring


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(a)

(b)

(c)

00.900.300.6SR A33.100.9

V0.12

S

R

VI

W6.1000.6A33.1 22 RIP

W31.500.3A33.1 22 RIP

W9.15W31.5W6.10 P

The total power delivered to any number of resistors in series is equal to

the power delivered to the equivalent resistance

20.6Series Wiring


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Personal electronic assistants.

20.7Paral lel Wirin g


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Paral le l wir ing means that the devices are connected in suc h a way

that the same voltage is appl ied acros s each device.



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PRV

RRV

RV

RVIII 111

2121

21

paral lel resistors: 321

1111

RRRRP

The total current from the voltage source is the sum of the currents

in the individual resistors.



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Example 10Main and Remote Stereo Speakers

Most receivers allow the user to connect to remote speakers in addition

to the main speakers. At the instant represented in the picture, the voltage

across the speakers is 6.00 V. Determine (a) the equivalent resistance

of the two speakers, (b) the total current supplied by the receiver, (c) the

current in each speaker, and (d) the power dissipated in each speaker.



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(a)

00.8

3

00.4

1

00.8

11

PR 67.2PR

(b) A25.267.2

V00.6rmsrms

PR

VI



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A750.000.8

V00.6rmsrms

R

VI(c) A50.1

00.4

V00.6rmsrms

R

VI

(d) W50.4V00.6A750.0rm srms VIP

W00.9V00.6A50.1rm srms VIP



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Conceptual Examp le 11A Three-Way Light Bulb

and Parallel Wiring

Within the bulb there are two separate filaments.

When one burns out, the bulb can produce only

one level of illumination, but not the highest.

Are the filaments connected in series orparallel?

20.8Circu its Wired Part ial ly in Series and Part ial ly in Paral lel


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Find(a) The total current supplied by

the battery

(b) Voltage between points A and

B in the circuit


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Exercise 5

Find the equivalent resistance of the circuit infigure below.

R = 15

20.9Internal Resistanc e


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Batteries and generators add some resistance to a circuit. This resistance

is called internal resistance.

The actual voltage between the terminals of a battery is known as theterminal vo ltage.

20.9Internal Resistanc e


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Example 12The Terminal Voltage of a Battery

The car battery has an emf of 12.0 V and an internal

resistance of 0.0100 . What is the terminal voltagewhen the current drawn from the battery is (a) 10.0 A

and (b) 100.0 A?

(a) V10.0010.0A0.10 IrV

11.9VV10.0V0.12

(b) V0.1010.0A0.100 IrV

11.0VV0.1V0.12

20.10Kirchhoffs Rules


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Thejunc t ion ru lestates that the total

current directed into a junction must

equal the total current directed out ofthe junction.



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The loop ru leexpresses conservation of energy in terms of the electricpotential and states that for a closed circuit loop, the total of all potential

rises is the same as the total of all potential drops.



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KIRCHHOFFS RULES

Junc t ion ru le. The sum of the magnitudes of the currents directed

into a junction equals the sum of the magnitudes of the currents directed

out of a junction.

Loop ru le. Around any closed circuit loop, the sum of the potential drops

equals the sum of the potential rises.



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Example 14Using Kirchhoffs Loop Rule

Determine the current in the circuit.



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risespotentialdropspotential

V240.8V0.612

II

A90.0I



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Example 15

Junction rule applied at B IA + IB = IH

Loop rule: BEFA IA (0.1) + 12.0 V = IB (0.1) +14.0 V

Loop rule: CDEB IB (0.1) + IH(1.2 ) = 12.0 V



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Reason ing Strategy

Applying Kirchhoffs Rules

1. Draw the current in each branch of the circuit. Choose any direction.

If your choice is incorrect, the value obtained for the current will turn out

to be a negative number.

2. Mark each resistor with a + at one end and a at the other end in a waythat is consistent with your choice for current direction in step 1. Outside a

battery, conventional current is always directed from a higher potential (the

end marked +) to a lower potential (the end marked -).

3. Apply the junction rule and the loop rule to the circuit, obtaining in the process

as many independent equations as there are unknown variables.

4. Solve these equations simultaneously for the unknown variables.


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Exercise 6

What is the emf of the battery in the circuit of figure below?

10.7 V

20.11The Measurement o f Current and Vol tage


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A dc galvanometer. The coil ofwire and pointer rotate when there

is a current in the wire.



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An ammeter must be inserted into

a circuit so that the current passes

directly through it.



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If a galvanometer with a full-scale

limit of 0.100 mA is to be used to

measure the current of 60.0 mA, ashunt resistance must be used so that

the excess current of 59.9 mA can

detour around the galvanometer coil.



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To measure the voltage between two pointsin a circuit, a voltmeter is connected between

the points.

20.12Capacitors in Series and Paral lel


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Paral lel capacito rs 321 CCCCP

VCCVCVCqqq 212121

20.12Capacitors in Series and Paral lel


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2121

21

11

CCq

C

q

C

qVVV

Series capacito rs 321

1111

CCCCS

20.13RC Circu its


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Capaci tor ch arging

RCt

o eqq 1

RC

time constant

20.13RC Circu its


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Capaci tor disch arging

RCt

oeqq

RC

time constant

20.13RC Circu its


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20.14Safety and the Phys io logical Effects of Cu rrent


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To reduce the danger inherent in using circuits, properelectr ical grou nd ing

is necessary.

Lecture3 (Chap 3)

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